Chords, triangles and segments

Take a sector $OAB$ and draw the straight line $AB$ joining the ends of the arc. That line is a chord, and it splits the sector into two pieces:

• the triangle $OAB$ (two sides are radii, so it's isosceles)
• the segment: the thin strip between the chord and the arc

Pizza version: the triangle is the cheesy middle, the segment is the crust you leave on the plate.

Throughout, $r$ is the radius and $\theta$ (the Greek letter "theta") is the angle $AOB$ at the centre, in radians.

You may know "area of a triangle $= \tfrac12 \times$ base $\times$ height". Picture this (the height line isn't drawn in the diagram): put $OB$ (length $r$) along the bottom as the base. The height is how far $A$ sits above the line $OB$. In the right-angled triangle formed by dropping a perpendicular from $A$ onto $OB$, the hypotenuse is $OA = r$ and the height is the side opposite the angle $\theta$, so

$\text{height} = r\sin\theta \qquad (\text{SOH: opposite} = \text{hypotenuse} \times \sin).$

So the area is $\tfrac12 \times r \times r\sin\theta$:

$\text{triangle } OAB = \tfrac12 r^2 \sin\theta.$

This is the general rule "area $= \tfrac12 ab\sin C$" (two sides times the sine of the angle between them, halved) with both sides equal to $r$. That rule works for any angle, obtuse ones included, so the formula holds for every sector, not just narrow ones.

The segment is the sector with the triangle cut out. The sector's area is $\tfrac12 r^2\theta$ (from the sectors lesson), so

$\text{segment} = \underbrace{\tfrac12 r^2\theta}_{\text{sector}} - \underbrace{\tfrac12 r^2\sin\theta}_{\text{triangle}} = \tfrac12 r^2(\theta - \sin\theta).$

The last step just takes out the common factor $\tfrac12 r^2$.

Look at the dashed line $OM$ in the diagram. Because $OA = OB$, the triangle is isosceles, and the line from $O$ to the midpoint $M$ of $AB$ meets $AB$ at a right angle and cuts the angle $\theta$ exactly in half.

Now use the right-angled triangle $OMB$:

• hypotenuse $OB = r$;
• angle at $O$ is $\tfrac{\theta}{2}$;
• $MB$ is opposite that angle, so $MB = r\sin\tfrac{\theta}{2}$.

The chord is two of these halves:

$AB = 2r\sin\tfrac{\theta}{2}.$

You can also get it from the cosine rule: $AB^2 = r^2 + r^2 - 2r^2\cos\theta$. Same answer, so pick whichever you trust.